Can elements of automorphism groups be expressed as some form of matrix? I was reading on automorphism groups and Galois theory and this idea came to mind:
Since by definition, automorphisms are isomorphisms, we have $\phi(a*b)=\phi(a)\cdot\phi(b)$ where $\phi$ is an element of the automorphism group $H$. The definition of $\phi$ reminds me of the definition of linear maps.
In the context of Galois extensions, if $H= Aut(F/\mathbb{Q})$(where $F/\mathbb{Q}$ is a Galois extension), does this mean that $\phi$ can be expressed as a matrix that transforms $F$ while fixing $\mathbb{Q}$ like its the "origin"?
If this is true, does this apply to any field extensions?
 A: If $L/F$ is a finite Galois extension, then the “finite” actually means that $L$ is a finite-dimensional vector space over $F$. Note that a field automorphism of $L$ over $F$ will actually also be a vector space automorphism of $L$ as a vector space over $F$ (it is additive, and for every $\alpha\in F$, $r\in L$, $\phi(\alpha r) = \phi(\alpha)\phi(r) = \alpha\phi(r)$, so it is also homogeneous). Thus, if you fix a basis for $L$ as an $F$-vector space, then the automorphism $\phi$ can be represented by a matrix.
However: not every vector space automorphism corresponds to a field automorphism (it need not be multiplicative for elements of $L$).
Also, the notion of Galois group and automorphism makes sense even if $L$ is not a finite extension, and in that case you can’t have a matrix representation of the automorphism in the usual sense.
A: In the context of Galois field, $Aut(F/\mathbb{Q})$ can be represented as matrix over the field $\mathbb{Q}$ if the dimension of $F$ over $\mathbb{Q}$ is finite as these are linear maps over the field $\mathbb{Q}$.
